B. One Bomb（水题）

B. One Bomb time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

You are given a description of a depot. It is a rectangular checkered field of n × m size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*").

You have one bomb. If you lay the bomb at the cell (x, y), then after triggering it will wipe out all walls in the row x and all walls in the column y.

You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall.

Input

The first line contains two positive integers n and m (1 ≤ n, m ≤ 1000) — the number of rows and columns in the depot field.

The next n lines contain m symbols "." and "*" each — the description of the field. j-th symbol in i-th of them stands for cell (i, j). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is occupied by a wall.

Output

If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes).

Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them.

Examples input 3 4 .*.. .... .*.. output YES 1 2 input 3 3 ..* .*. *.. output NO input 6 5 ..*.. ..*.. ***** ..*.. ..*.. ..*.. output YES 3 3

其实是一道很简单的暴力水题，但是也要注意一定的方式方法，灵活运用标记数组

如果暴力判断的话会超时，而是选择标记数组，将x对应的炸弹值和y对应的炸弹值相加若正好等于总的炸弹值则为结果，但是注意有没有x数组和y数组都标记的位置

#include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath> using namespace std; const int maxx=1005; int x,y; char mp[maxx][maxx]; int xx[maxx],yy[maxx]; int main() { scanf("%d %d",&x,&y); for(int i=0;i<x;i++) { scanf("%s",mp[i]); } int sum=0; for(int i=0;i<x;i++) { for(int j=0;j<y;j++) { if(mp[i][j]=='*') { xx[i]++;yy[j]++; sum++; } } } int flag=0; for(int i=0;i<x;i++) { for(int j=0;j<y;j++) { if((mp[i][j]=='*'&&xx[i]+yy[j]==sum+1)||(mp[i][j]!='*'&&xx[i]+yy[j]==sum)) { printf("YES\n%d %d\n",i+1,j+1); flag=1; break; } } if(flag==1) { break; } } if(flag==0) { printf("NO\n"); } return 0; }